1. Key areas
The relationship between the wavelength, distance between the sources, distance from the sources and the spacing between maxima or minima.
The relationship between grating spacing, wavelength and angle to the maxima.

2. What we will do today:
State the equation used to describe the relationship between grating spacing, wavelength and angle to the maxima.
Carry out calculations on the above.
State the relationship between the wavelength, distance between the sources, distance from the sources and the spacing between maxima or minima.

3. The Diffraction Grating

4. The Grating and Monochromatic Light
A grating consists of many equally spaced slits positioned extremely close together (e.g. 300 lines per mm).
Light is diffracted through each slit and interference takes place in a similar fashion to the double slit.

5. The advantage of the grating is that much more light is transmitted through and a clearer interference pattern is seen.

6. Zero Order θ
Laser
Monochromatic Light
Source
(Light of one frequency)
Grating
e.g. 300 lines/mm
Screen

7. The Grating Equation For a grating: n = order of the maximum
nλ = d sinθ
n = order of the maximum
λ = wavelength of light
d = separation of slits
θ = angle from zero order to nth maximum
λ and d must be in same units!!!

8. Separation of Slits
d is the distance between the slits on the grating.
It is usually very small, e.g m.
You can find d if you are given N, the number of slits (or lines) per metre on the grating.
d = 1 / N

9. Example:
N = 500 lines per mm
= 500 x 1000 lines per m
= 5 x 105 lines per m
d = 1 / N = 1 / (5 x 105) = 2 x 10-6m

10. If the grating equation is rearranged to sinθ = nλ / d, then it can be seen that to increase θ (the separation of the maxima) you can:
Increase the wavelength (move from blue towards red light).
Decrease the slit separation (have more lines per mm).
Move the screen further away.

11. Example
A diffraction grating with 300 lines per mm is used to produce an interference pattern. The 2nd order maximum is obtained at a diffracted angle of 19º.
Calculate the wavelength of the light.

12. Solution Using the formula nλ = d sinθ: d = 1/300 mm = 3.33 x 10-3 mm
θ = 19º
nλ = d sinθ
2 x λ = 3.33 x 10-6 sin19º
λ = 5.4 x 10-7 m
= 540 nm

13. Questions
Class jotter
Qu’s:

14. Past Paper Questions 2006 – Qu: 27(b) 2005 – Qu: 28(b)

15. (b)

16. (b)

17. (b)

18. Revised Higher
Past Paper Questions

19. (b)

20. (b)